add math~~M.I.

Let P(n) be the statement "n (n+1) (n+2) is divisible by 3", where n is a positive integer.

When n=1,
1(1+1)(1+2) = 6, which is divisible by 3
Therefore, P(1) is true.

Assume P(m) is true,
i.e. m(m+1)(m+2) = 3M, where M is an positive integer.

When n=m+1,
(m+1)(m+1+1)(m+1+2)
= m(m+1)(m+2) + 3(m+1)(m+2)
= 3M + 3(m+1)(m+2)
= 3[M + (m+1)(m+2)]
Therefore, P(m+1) is true.

By the principle of mathematical induction, P(n) is true for all positive integers n.

let S(n) be the statement 'n (n+1) (n+2) is divisible by 3 '
let f(n)=n (n+1)(n+2)
when n=1
f(1)=1(2)(3)
=6
S(1) is true

assume S(k) is true
k(k+1)(k+2)=3m where m is some positive integer

when n=k+1
f(k+1)=(k+1)(k+2)(k+3)
=(k+1)(k+2)(k)+3(k+1)(k+2)
=3m+3(k+1)(k+2)
=3[m+(k+1)(k+2)]

S(k+1) is true

By M.I., S(n) is true for all positive integer n